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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
5. Prove that 3^(1/2) x 3^(1/4) x 3^(1/8)......=3 |
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| 2. |
(छु0ड tanA 1.+ goe A1-cosA |
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| 3. |
Solve for x:3\left(\frac{3 x-1}{2 x+3}\right)-2\left(\frac{2 x+3}{3 x-1}\right)=5 ; x \neq \frac{1}{3},-\frac{3}{2} |
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| 4. |
11-tane(ख) सिद्ध कीजिए : 1+|11-cote= sec . |
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Answer» , like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like like like Like |
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| 5. |
If sece + tane p, then find the value of cosece |
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Answer» Like if you find it useful |
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| 6. |
Find the zeroes of the given poly(i) p(x) = 3x (ii) p(x) = x^2 +ill) p(x) = ( x+2) ( x+ 3) (iv) p(x) = x^4-16 |
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| 7. |
2 If tane-u, find the value of sec θ.b' |
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Answer» brother u r completely wrong... I thought that sm1 will help and u did tq but it's wrong ..I have ans in my guide but I don't some steps this is the answer now if u know steps to get this ans then plz help bro |
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| 8. |
30. Prove the trigonometric identity:-cote+1-tanetan θ1+ sec 0. cosec01 cote |
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Answer» 1 |
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| 9. |
lf5x-sec eared,' tane,find the value of 5(x2-3) |
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| 10. |
If tane 1-e then show that sec+tan3e coseco (2 - e)32 |
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| 11. |
sin 2x + costs-I - PFind value of P. |
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Answer» the value of p is 0 1-1=0 |
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| 12. |
find the value of p, if p+6=3 |
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| 13. |
If cosec θ + cot θ-p, then prove that cos θ---p + 12 |
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| 14. |
\operatorname { sec } \theta + \operatorname { tan } \theta = p , \text { prove that } \operatorname { sin } \theta = \frac { p ^ { 2 } - 1 } { p ^ { 2 } + 1 } |
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| 15. |
(i) tanA +sinA secA+1Âť' | tan A âsin A e ) |
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| 16. |
( x + y ) ^ { - 1 } \cdot ( x ^ { - 1 } + y ^ { - 1 } ) = x ^ { p } y ^ { q } , \text { prove that } p + q + 2 = 0 |
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| 17. |
(cosecA - sinA ) (secA - cosA ) = 1/tanA + cotA |
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| 18. |
(secA+tanA)(1-sinA) |
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Answer» tnx it's right |
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| 19. |
x236 value of p(3) + p(0) + p(l).5x + 2, then find the2a 18. If pix)19. If α, β and γ be the zeroes of the cubicoes( 4x2 + 5x + 2,βγ+γα.polynomial p(x) 3x3 +then find the value of oB + |
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| 20. |
if secA + tanA = Pfind value of cosecA |
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Answer» this question has already been answered answer is not showing |
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| 21. |
If p(x) = x+5 then find the value of p(x) + p(-x) |
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Answer» p(x)+p(-x) =x+5-x+5=10 p(x)=x+5......eq 1p(_x)=x_5.........eq 2on adding eq 1 and 2 we get, 0 p(x)=x+5;q(-x)= -x+5; then p(x)+q(-x)=x+5+(-x+5)=x+5-x+5=10 answer will be zero bcoz the sign changes in p(-x) the answer is 0 because p(x) + p(-x) is equal to 0 P(x) + P(-x), Here, P(x) =x+5so, x+5+[-(x+5) ]=x+5-x-5=0 If P(x)=x+5 then P(-x) =-x+5 Therefore, P(x)+P(-x) =x+5-x+5= 10. just put -ve in x term i.e. we get 10 after adding x+5 and -x+5 P(x) =x+5Then,P(-x)=-(x+5)P(-x)=-x-5Now,P(x)+p(-x)=x+5-x-5=0 right answer is 10.if sine change for that only the sine change of the value of the x not the value of total function p(x)=x+5 p(-x)=-x+5 Then p(x)+p(-x) = x+5-x+5= 10 if p(x)=(x+5),. then p(-x)=(-x-5)now add p(x)&p(-x)we getx+5-x-5=0 p(x)+p(-x)p(x)-p(x)px-px0 p(x)=x+5 p(-x)=-x+5adding both we get 10 the correct answer is zero X+5 -x+5=10 so this is the answer of this question p(x)+p(-x)p(-5)+p(+5)2p if P(x)= x+5 P(x) +P(-x) x+5+x-5value is 2x 2p are the right answer p(x)+p(-x)=x+5-x+5=10 p(x)=x+5x=-5 y(x)=p(x)+p(-x) =p(-5)+p(+5)so -5 or +5 is cancel so that the answer is=20 p(X)+p(-X) = (X+5)+(-(X+5)=0 p(-x)+p(x)=-x+5+x+5=10 x+5-x+5 = 10 given that p(x) = x+5so just put the value in the equation given (x+5) + (-x+5) = 10 p(x)+p(-x)=x+5-x-5=0 the correct answer is 10 P (x)=x+5,so P (-x)=-x+5,now P (x)+P(-x)=x+5+(-x+5)=x+5-x+5=10 the answer is 0 The anser to the above question is 10 p(x)=-5x=-5p(-5)+p(-(-5))-5p+5p=0 p(x)+p(-x )= X+5-x-5=0 -5 haiiiiiiiiiiiiiiiiiiiiii p(x)=x+5p(-x) =-(x+5) = -x-5 p(x)+p(-x) =x+5+(-x-5) =x+5-x-5 =0 if p(x) =x+5p(-x) =-x+5p(x)+p(-x) =x+5-x+5=10 (x+5)+(-x+5) = 5+5 = 10 p(x)=x+5then p(-x)=-x+5p(x)+p(-x)=x+5-x+5 =10 10 is the correct answer for this x+5=0x= -5 x-5=0x=5 p(5)+p(-(-5))5+5=10 p(-5)+p(5)=-5p+5p = 0 your answer is zero only zero x+5-x-5=0 . 0 is the answer the answer of this question will be 10 The answer will be 10. Put - x instead of x in 2nd term. P(x) +p x+5-x+5=10 so the answer is 10 given, p(x) = x + 5then, p(-x)= -x + 5p(x)+p(-x)= x + 5 - x + 5 = 10 p(x)=x+5p(-x)It can also be written as -pxTherefore p(-x)=-(x+5)=-x-5Therefore p(x)+p(-x)=x+5+(-x-5)=x+5-x-5=0 in above problem if p(x)=x+5 then p(-x)=-x+5 and hence p(x)+p(-x)=x-x+10=10 Bhai aapk x ki jagah -5 rakhi aapka answer aa jayega 10 is the absolutely correct answer P(X) = X + 5 & P(-X) = -X+5 so adding it we get 10 ans |
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| 22. |
louar iulll the puLHt BI3, p and C(p, 5), then find the valueof p.5. If the quadratic equation px2-2/5px +15 0 has two equal roota, then find thevalue of p. |
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Answer» Equal rootsmeans D= 0b^2= 4ac Compare px² -2√5 px + 15 = 0 with ax² + bx + c = 0 a = p , b = -2√5 , c = 15 We know that , If the roots of the quadratic equation are equal , then it's discriminant (D) equals to zero. D = 0 b² - 4ac = 0 ( -2√5 p )² - 4×p ×15 = 0 20p² - 60p = 0 20p( p - 3)= 0 Therefore , 20p = 0 or ( p- 3 ) = 0 p = 0 or p = 3 I hope this helps you |
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| 23. |
If p= 4xy/x+y then find the value of p+2x/p-2x + p+2y/p-2y |
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| 24. |
If (0,P) is solution of equation 5x_3y=0.then find value P. |
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Answer» (0,P) is a solution of 5x - 3y = 05×0 - 3×P = 03P = 0P = 0 |
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| 25. |
es ol TA +SK +/, find the value of 1/ι + 1/β10.Show that tana+ tan (90-0)-sec0. sec (90- P) |
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| 26. |
lete the multiplication matrix by finding the missing numbersC, D. Then the value of A + B + C + D will be(b) 2805(d) 18055x|22| 1135 | AB50 |
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| 27. |
if secA+tanA=p then prove p^-1/p^+1=sinA |
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Answer» secA + tanA = p1/cosA + sinA/cosA = p(1 + sinA)/cosA = p(1 + sinA)^2/cosA^2 = p^2[squaring both sides](1 + sinA)^2/(1 - sinA^2) = p^2[ since cosA^2 + sinA^2 = 1](1 + sinA)^2/(1 + sinA)(1 - sinA) = p^2(1 + sinA)/(1 - sinA) = p^2[the common part getting cancelled]2/2sinA = (p^2 + 1)/(p^2 - 1)[By componendo and dividendo we get]1/sinA = (p^2 + 1)/(p^2 - 1) ThereforesinA = (p^2 - 1)/(p^2 + 1)[Answer] |
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| 28. |
ll At the same time, in how much time willompletely?A carpenter can do a piece of work in 5 days, but with the help of his son, he can do it in 3 days. In how manydays can the son do the work alone?1. |
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| 29. |
Q.17 If A+B 90 prove thattanA tanB + tanA cotB sin2 Bcos2AtanAsinA secA |
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Answer» Given, LHS = root tan a tan b + tan a cot b/sin a sec b - sin^2 b/cos^2 a = root tan a tan(90-a) + tan a cot(90-a)/sin a sec(90-a) - sin^2(90-a)/cos^2 a = root tan a.cot a + tan a.tana/sin a.cosec a - cos^2 a/cos^2 a = root 1+tan^2 a/1 - 1 = root tan^2 a = tan a. |
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| 30. |
tanA/1-cotA + cotA/1-tanA |
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| 31. |
) The sum of the digits of a two-digit number is 11. The number got byinterchanging the digitsis 27 more than the original number. What is thisnumber? |
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| 32. |
tanA/1+secA-tanA/1-secA=2cosesA |
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Answer» tanA / secA - 1 + tanA / secA + 1 sinA/cosA / 1/ cosA - 1 + sinA/cosA / 1/cosA + 1 on simplifying sinA / 1 - cosA + sinA / 1 + cosA sinA + sinA.cosA + sinA - sinA.cosA / 1 - cos2A 2sinA / sin2A 2 / sinA 2cosecA Like my answer if you find it useful! |
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| 33. |
h[ & Call LetE/18)], |
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Answer» [5 ( 8^1/3 + 27^1/3)^3]^4 = [5 ( 2 + 3)^3]^4 = [5 * (5)^3]^4 = (5^4)^4 = 5^16 |
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| 34. |
Principal5000, rate9% pa. and time146 days. |
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Answer» P= Rs5000, R=9% and T=146 days....=146/365S.I.=P*R*T/100=5000*9*146/100*365=180Rs=5000+180=Rs 5180 |
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| 35. |
हा का अन्तर 14 तथा उसके वर्गों का अन्तर 448 है। संख्याएं, ज्ञात |
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| 36. |
8, Principal =5000, amount =6450 and rate =12% pa |
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| 37. |
5/Principal =5000, rate9% pa. and time146 days. |
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Answer» Time = 146 days , in year it is = 146/365 = 0.4 years now I = P*R*T/100 = 5000*9*0.4/100 = 180 so, A = P+I = 5000+180 = 5180 thanks Devan |
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| 38. |
2. SIMPLE INTEREST(2) Principal, 5000, Rate of interest= 10.5% and Period= 3 years |
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Answer» P = 5000, R = 10.5, T = 3 yrsSI = P*R*T/100 = 5000*10.5*3/100 = 15*10.5 = 157.50 Simple Interest is Rs 157.5 thank you |
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| 39. |
The difference between two numbers is 14 and the difference between their squares is 448. Find the numbers. |
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| 40. |
watch sold for? 448 gives a profit of only 12%. Find the profit % if the selling price had been512. |
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Answer» Selling price of the watch = Rs. 448 Profit earned = 12 % Cost price = ? Let the cost price of the watch be Rs. 'x' Then according to the question. ⇒ x/1 + (x*12)/100 = 448 Taking L.C.M. of the denominators and then solving it. ⇒ (100x + 12x)/100 = 448 ⇒ 112x = 448*100 ⇒ x = 44800/112 ⇒ x = Rs. 400 So, the cost price of the watch is Rs. 400 Now, we have to find the profit % if the selling price is Rs. 512 Selling price - Cost price = Profit Profit = 512 - 400 Profit = Rs. 112 Profit % = (Profit*100)/Cost Price ⇒ (112*100)/400 ⇒ 11200/400 ⇒ 28 % So, if the selling price of the watch is Rs. 512, then the profit is 28 % |
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| 41. |
10. The difference between two numbers is 14 and the difference betweentheir squares is 448. Find the numbers. |
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| 42. |
\left(\frac{2 x+1}{x+1}\right)^{4}-13\left(\frac{2 x+1}{x+1}\right)^{2}+36=0 |
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| 43. |
f(x)=\left\{\begin{array}{l}{\frac{\sqrt{1+k x}-\sqrt{1-k x}}{x}, \text { if }-1 \leq x<0} \\ {\frac{2 x+1}{x-1}, \quad, \text { if } 0 \leq x<1} \\ {\frac{2 x+1}{x-1}, \quad, \text { if } 0 \leq x<1}\end{array}\right. |
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| 44. |
GM. and A.M. between two numbers are 8 and 17. Find the numbers. |
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| 45. |
If(x-x)-4, find the value of(i) (x2+12)(ii) (X4)If] x-스 | = 4, find the value of(i) | x2 +, |
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| 46. |
66. (secA+tanA -1)(secA-tanA +1) = 2tan A |
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Answer» secA+(tanA-1)}{secA-(tanA-1)}=(secA)² - (tanA-1)² =sec²A - tan²A - 1 + 2tanA =1 - 1 + 2tanA =2tanA |
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| 47. |
(1) 1/seca-tana – 1/cosa = 1/cosa-1/seca+tana (ii) 1/secA +1/secA+1= 2cosecAcot A |
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Answer» LHS=1/seca-tana – 1/cosa = cosa /(1- sina) – 1/cosa = (cos^2a – (1 – sina))/(1 – sina)(cosa) = (sina – sin^2a)/(1 – sina)(cosa) = (sina)(1 – sina)/(1 – sina)(cosa) = tanaRHS=1/cosa – 1/(seca+tana) = (1/cosa) –( cosa/(sina+1)) = (sina +1 – cos^2a)/(cosa)(sina+1) = (sina +1 +sin^2a – 1)/(sina+1)(cosa) = sina/cosa = tana [1/(SecA-1)]+[1/(SecA+1)] =[1*(SecA+1) + 1*(SecA-1)]/[(SecA-1)(SecA+1)] =[SecA + 1 + SecA - 1]/[Sec^2A - 1^2] =[SecA + 1 + SecA - 1]/[Tan2A][Using: Sec2Θ-Tan2Θ=1 => Sec2Θ-1=Tan2Θ] =[SecA +1+ SecA -1]/[Tan^2A] =[2SecA]/[Tan^2A] =[2(1/CosA)]/[(Sin^2A/Cos^2A)] =[(2*1)/CosA]*[Cos^2A/Sin^2A] = [2/CosA] * [Cos^2A/Sin^2A] = [2] * [CosA/Sin^2A] = 2*(CosA/SinA)*(1/SinA) = 2CotACosecA |
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| 48. |
Neetug nseight decreasad ë8M 60 heto y5Rg.findcentaeticsc |
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Answer» Original weight=60Decreased weight=45percentage decrease=((60-45)/60) x 100=25% hgfi |
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| 49. |
X+3=4,find value of x? |
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Answer» X=4-3X=1is the right answer |
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| 50. |
2. Find the sum of the following scries0) 1+4+7+10+... to 40 terms |
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